How to prove inequality including natural log? Let $0<p<1$ and $0<q<1$ be real numbers and $0 < p + q < 1$.
How can I prove the following inequality is correct or not?
$$q > \frac{\ln(\frac{p}{1-q})}{\ln(\frac{p}{1-q}) + \ln(\frac{q}{1-p})}$$
$\ln$ is natural logarithm.
 A: Recall the definition of Relative Entropy. In this case:
$D(q||(1-p))=\sum\limits_{x\in\mathcal{X}} q(x)\ln\left(\frac{q(x)}{1-p(x)}\right)$.
In the binary case:
$D(q||(1-p)) = q\ln(q)-q\ln(1-p)+(1-q)\ln(1-q)-(1-q)\ln(p)\geq 0$
So,
$D(q||(1-p)) = q\ln(\frac{q}{1-p})+q\ln(\frac{p}{1-q})-\ln(\frac{p}{1-q}) \geq 0$
EDIT
Use the fact that relative entropy is non-negative. The equality holds iff the two distributions are identical. The last condition actually means that two distributions 1-p and q are not identical, so the relative entropy in this case is positive.
Correction
Since $p+q<1$, then $p<1-q$ and $q<1-p$, so $\ln(\frac{p}{1-q})$ and $\ln(\frac{q}{1-p})$ are both negative, thus
$q\ln(\frac{q}{1-p})+q\ln(\frac{p}{1-q})-\ln(\frac{p}{1-q}) > 0$
$\Rightarrow q\left[ \ln(\frac{q}{1-p})+\ln(\frac{p}{1-q})\right] > \ln(\frac{p}{1-q}) $
$\Rightarrow q < \dfrac{\ln(\frac{p}{1-q})}{\ln(\frac{q}{1-p})+\ln(\frac{p}{1-q})} $
A: Since
$$0\lt p\lt 1$$
$$0\lt q\lt 1$$
$$0 \lt p + q \lt 1$$
Then
$$\frac{p}{1-q}\lt 1\Rightarrow\ln\left(\frac{p}{1-q}\right)\lt 0$$
$$\frac{q}{1-p}\lt 1\Rightarrow\ln\left(\frac{q}{1-p}\right)\lt 0$$
So now we have
$$q\gt\frac{\ln\left(\frac{p}{1-q}\right)}{\ln\left(\frac{p}{1-q}\right) + \ln\left(\frac{q}{1-p}\right)}$$
$$\frac{q}{1-q}\ln\left(\frac{q}{1-p}\right)\lt\ln\left(\frac{p}{1-q}\right)$$
Here we can use the fact that for $0\lt x\lt 1$, we have
$$\frac{x-1}{x}\lt\ln(x)\lt\frac{(x-1)(x+5)}{2(2x+1)}$$
Which yields
$$\ln\left(\frac{p}{1-q}\right)\lt\frac{(p+q-1)(p-5q+5)}{2(2p-q+1)(1-q)}$$
And
$$\frac{p+q-1}{1-q}\lt\frac{q}{1-q}\ln\left(\frac{q}{1-p}\right)$$
If the original inequality were true, one would expect the upper bound to be greater than the lower bound. However
$$\frac{p+q-1}{1-q}\lt\frac{(p+q-1)(p-5q+5)}{2(2p-q+1)(1-q)}$$
$$2(2p-q+1)\gt p-5q+5$$
$$p+q\gt 1$$
Therefore
$$\frac{q}{1-q}\ln\left(\frac{q}{1-p}\right)\not\lt\ln\left(\frac{p}{1-q}\right)$$
